Physics - Modern Physics Question with Solution | TestHub

Using Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K=\frac{hc}{\lambda }-\varphi$, where $\varphi$ is the work function of the material.

For the first case with wavelength $\lambda$:
$K_1=\frac{hc}{\lambda }-\varphi$

For the second case with wavelength $\frac{\lambda }{3}$:
$K_2=\frac{hc}{\frac{\lambda }{3}}-\varphi =\frac{3hc}{\lambda }-\varphi$

It is given that the maximum kinetic energy in the second case is $4$ times that in the first case:
$K_2=4K_1$

Substituting the expressions for $K_1$ and $K_2$:
$\frac{3hc}{\lambda }-\varphi =4\left(\frac{hc}{\lambda }-\varphi \right)$

$\frac{3hc}{\lambda }-\varphi =\frac{4hc}{\lambda }-4\varphi$

$4\varphi -\varphi =\frac{4hc}{\lambda }-\frac{3hc}{\lambda }$

$3\varphi =\frac{hc}{\lambda }$

$\varphi =\frac{hc}{3\lambda }$